2016 Paper 2 Q10
Year: 2016
Paper: 2
Question Number: 10
Course: UFM Mechanics
Section: Centre of Mass 1
Difficulty: 1600.0
Banger: 1516.0
Problem
A thin uniform wire is bent into the shape of an isosceles triangle \(ABC\), where \(AB\) and \(AC\) are of equal length and the angle at \(A\) is \(2\theta\).
The triangle \(ABC\) hangs on a small rough horizontal peg with the side \(BC\) resting on the peg. The coefficient of friction between the wire and the peg is \(\mu\). The plane containing \(ABC\) is vertical. Show that the triangle can rest in equilibrium with the peg in contact with any point on \(BC\) provided
\[
\mu \ge 2\tan\theta(1+\sin\theta)
\,.
\]
Solution
Examiner's report
— 2016 STEP 2, Question 10
Below Average
Fewer than 400 attempts
From the paper introduction
As in previous years the Pure questions were the most popular of the paper with questions 1, 3 and 7 the most popular of these. The least popular questions of the paper were questions 10, 11, 12 and 13 with fewer than 400 attempts for each of them. There were many examples of solutions in this paper that were insufficiently well explained, given that the answer to be reached had been provided in the question.
Source: Cambridge STEP 2016 Examiner's Report
· 2016-full.pdf
Rating Information
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
Show LaTeX source
Problem source
A thin uniform wire is bent into the shape of an isosceles triangle $ABC$, where $AB$ and $AC$ are of equal length and the angle at $A$ is $2\theta$.
The triangle $ABC$ hangs on a small rough horizontal peg with the side $BC$ resting on the peg. The coefficient of friction between the wire and the peg is $\mu$. The plane containing $ABC$ is vertical. Show that the triangle can rest in equilibrium with the peg in contact with any point on $BC$ provided
\[
\mu \ge 2\tan\theta(1+\sin\theta)
\,.
\]
Solution source
\begin{center}
\begin{tikzpicture}[scale=2]
\coordinate (A) at (0,2);
\coordinate (B) at (-1,0);
\coordinate (C) at (1,0);
\coordinate (O) at (0,0);
\coordinate (G) at (0, {2/3});
\filldraw ($(A)!0.5!(B)$) circle (0.8pt) node [left] {$c$};
\filldraw ($(C)!0.5!(B)$) circle (0.8pt) node [below] {$a$};
\filldraw ($(A)!0.5!(C)$) circle (0.8pt) node [right] {$b$};
\draw (A) -- (B) -- (C) -- cycle;
\draw[dashed] (A) -- (O);
\node[above] at (A) {$A$};
\node[left] at (B) {$B$};
\node[right] at (C) {$C$};
\pic [draw, angle radius=1cm, angle eccentricity=1.25, "$\theta$"] {angle = B--A--O};
\end{tikzpicture}
\end{center}
Clearly the centre of mass will lie on the perpendicular from $A$. We can also consider each side's wire as equivalent to a point mass at the centre of the side with mass proportional to the length of the side. Recalling that $b = c$ (the triangle is isoceles we must have (for the $y$-coordinate
\begin{align*}
&& a \cdot 0 + b \cdot \frac12 b \cos \theta + c \cdot \frac12 c \cos \theta &= (a+b+c) \overline{y} \\
\Rightarrow && b^2 \cos \theta &= (2b + 2b\sin \theta) \overline{y} \\
\Rightarrow && \overline{y} &= \frac{b \cos \theta}{2(1+\sin \theta)}
\end{align*}
\begin{center}
\begin{tikzpicture}[scale=2]
\begin{scope}[rotate=160]
\coordinate (A) at (0,2);
\coordinate (B) at (-1,0);
\coordinate (C) at (1,0);
\coordinate (O) at (0,0);
\coordinate (G) at (0, {2/3});
\coordinate (X) at ({2/3*sin(20)},0.04);
\coordinate (Xx) at ({2/3*sin(20)},0.0);
\coordinate (Y) at ({2/3*sin(20)},-0.5);
\coordinate (Z) at ({2/3*sin(20)+0.5},0);
\end{scope}
\filldraw (G) circle (1.2pt) node [right] {$G$};
\draw (A) -- (B) -- (C) -- cycle;
\draw (X) circle (1.2pt);
\draw[dashed] (A) -- (O);
\node[below] at (A) {$A$};
\node[right] at (B) {$B$};
\node[left] at (C) {$C$};
\draw[-latex, ultra thick, blue] (Xx) -- (Y) node[above] {$R$};
\draw[-latex, ultra thick, blue] (Xx) -- (Z) node[left] {$F$};
\draw[-latex, ultra thick, blue] (G) -- ++ (0,-0.5) node[below] {$mg$};
\pic [draw, angle radius=1cm, angle eccentricity=1.25, "$\theta$"] {angle = B--A--O};
\end{tikzpicture}
\end{center}
\begin{align*}
\text{N2}(\nearrow): && R - mg \cos \phi &= 0 \\
\text{N2}(\nwarrow): && F -mg \sin \phi &= 0 \\
\Rightarrow && F &\leq \mu R \\
\Rightarrow && \sin \phi &\leq \mu \cos \phi \\
\Rightarrow && \tan \phi &\leq \mu
\end{align*}
When the peg is at $C$
\begin{align*}
\tan \phi &= \frac{CM}{MG} \\
&= \frac{b\sin \theta}{\frac{b \cos \theta}{2(1+\sin \theta)}} \\
&= 2 \tan \theta(1+\sin \theta)
\end{align*}
Therefore $2 \tan \theta(1+\sin \theta) \leq \mu$ as required.
This question received a number of very good, and often concise, answers. However, there was a significant number of candidates who did not calculate the centre of mass, or mistakenly assumed that the formula for the centre of mass for an equilateral triangle could be applied. Many candidates chose to consider the limiting case first and then deduce the inequality in the final step, but did not justify the direction of the inequality clearly. There were a number of cases where the required angles where not calculated correctly when resolving the forces.